Answer: 653.011Mpc
Explanation:
Hubble deduced that the farther the galaxy is, the more redshifted it is in its spectrum, and noted that all galaxies are "moving away from each other with a speed that increases with distance", and enunciated the now called Hubble–Lemaître Law.
This is mathematically expressed as:
[tex]V=H_{o}D[/tex] (1)
Where:
[tex]V[/tex] is the approximate recession velocity of the galaxy
[tex]H_{o}=70 km/s/Mpc[/tex] is the current Hubble constant
[tex]D[/tex] is the galaxy's distance
On the other hand, the equation for the Doppler shift is:
[tex]\frac{\Delta \lambda}{\lambda_{o}}=\frac{V}{c}[/tex] (2)
Where:
[tex]\lambda_{o}=656.3nm=656.3(10)^{-9}m[/tex] is the wavelength for the Ha line of the galaxy observed at rest
[tex]\Delta \lambda=\lambda_{1}-\lambda_{o}[/tex] is the variation between the measured wavelength for the Ha emission line in the spectrum of this galaxy ([tex]\lambda_{1}=756.3nm=756.3(10)^{-9}m[/tex] in this case) and the wavelength for the same Ha line observed at rest
[tex]c=3(10)^{8}m/s[/tex] is the speed of light
Rewriting (2):
[tex]\frac{\lambda_{1}-\lambda_{o}}{\lambda_{o}}=\frac{V}{c}[/tex] (3)
Isolating [tex]V[/tex]:
[tex]V=\frac{(\lambda_{1}-\lambda_{o})c}{\lambda_{o}}[/tex] (4)
Finding [tex]V[/tex]:
[tex]V=\frac{(756.3(10)^{-9}m-656.3(10)^{-9}m)3(10)^{8}m/s}{656.3(10)^{-9}m}[/tex] (5)
[tex]V=45710802.99m/s=45710.80299km/s[/tex] (6)
Substituting [tex]V[/tex] in (1):
[tex]45710.80299km/s=(70km/s/Mpc)D[/tex] (7)
Finding [tex]D[/tex]:
[tex]D=\frac{V}{H_{o}}=\frac{45710.80299km/s}{70km/s/Mpc}[/tex] (8)
Finally:
[tex]D=653.011Mpc[/tex] (8) This is the galaxy's distance
